How do I evaluate #int \ ln(2x+1) \ dx#?
1 Answer
Apr 10, 2018
Explanation:
Note that:
#d/dx((2x+1) ln(2x+1)) = 2ln(2x+1)+(2(2x+1))/(2x+1)#
#color(white)(d/dx((2x+1) ln(2x+1))) = 2ln(2x+1)+2#
So:
#d/dx(1/2(2x+1)ln(2x+1)-x) = ln(2x+1)#
So:
#int \ ln(2x+1) \ dx = 1/2(2x+1)ln(2x+1)-x+C#
